Asymptotic expansion for $xe^{1/x}=e^{\lambda}$

Question

I need to find a 3-term expansion as $\lambda\rightarrow\infty$ for each of the solutions to this transcendental equation. Plotting $xe^{1/x}$ I assume that there are 2 solutions, one around 0 and the other at infinity: plot

I've played around a little thinking I might be able to use an iterative method but that didn't work: $ln(xe^{1/x})=ln(e^{\lambda})$ $\rightarrow$ $x_{n+1}=\frac{1}{\lambda}-\frac{1}{ln(x_{n})}$

Any ideas? thanks

Edit: Thanks @ClaudeLeibovici, I've gone down that path before but it didn't seem to work (for me). I tried solving this again using dominant balance, not sure about the final expressions:

1. $\frac{1}{x}\sim\lambda\rightarrow x\ll1\rightarrow\ln\left(x\right)\ll1$

   Is one valid balance which leads to: $x\sim\frac{1}{\lambda}+x_{0}\quad x_{0}\ll\frac{1}{\lambda}$

   $\ln(\frac{1}{\lambda}+x_{0})+\frac{1}{\frac{1}{\lambda}+x_{0}}=\lambda$

   $\ln(\frac{1}{\lambda}\left(1+\lambda x_{0}\right))+\frac{1}{\frac{1}{\lambda}\left(1+\lambda x_{0}\right)}=\lambda$

   $\underset{\ll1}{\underbrace{\ln(\frac{1}{\lambda})}}+\ln\left(1+\underset{\ll1}{\underbrace{\lambda x_{0}}}\right)+\frac{\lambda}{\left(1+\lambda x_{0}\right)}=\lambda$

   $...x_{0}=\frac{1}{\lambda^{2}}$

   Using Taylor expansions and neglecting smaller terms.

2. The other valid dominant balance is: $\ln\left(x\right)\sim\lambda\rightarrow x=ord\left(\lambda\right)\rightarrow\frac{1}{x}\ll1$

   which leads to: $\ln\left(x\right)\sim\lambda\rightarrow x\sim e^{\lambda}+x_{1}\quad x_{1}\ll e^{\lambda}$

   $ln(e^{\lambda}+x_{1})+\frac{1}{e^{\lambda}+x_{1}}=\lambda$

   $ln(e^{\lambda}\left(1+\frac{x_{1}}{e^{\lambda}}\right))+\frac{e^{\lambda}}{1+\frac{x_{1}}{e^{\lambda}}}=\lambda$

   $...x_{1}=\frac{e^{\lambda}}{\lambda}$

Does that seem correct?

Answer 1

Hint

First, plot the function $$f(x)=x\,e^{\frac{1}{x}} $$ and notice how close to linearity it is. So, for solving $f(x)=k$ a first order approxiamtion is $x_0=k$.

Now, the rhs $(k=e^\lambda)$ can be very large. So, your idea of using logatithms is good, that is to say that now you will try to find the zero of $$g(x)=\log \left(x\,e^{\frac{1}{x}} \right)-\lambda=\frac{1}{x}+\log(x)-\lambda$$ and use $x_0=e^\lambda$.

Now, compute $x_1$ as the first iterate of Newton method.